90/VB02. problem / Math - Solved Math Problems

MR-90 / 92a. problem

Rationalize the denominator of the fraction:

$\frac{6}{\sqrt[3]{2}}$

\frac{6}{\sqrt[3]{2}}

$\frac{6}{\sqrt[3]{2}}$

$= \frac{6}{\sqrt[3]{2}} \cdot \frac{\sqrt[3]{2^{2}}}{\sqrt[3]{2^{2}}}$

= \frac{6 \cdot \sqrt[3]{2^{2}}}{\sqrt[3]{2^{3}}}

= \frac{6 \cdot \sqrt[3]{2^{2}}}{2}

$= \frac{3 \cdot 2 \cdot \sqrt[3]{2^{2}}}{2}$

$= 3 \sqrt[3]{4}$

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